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104

Dynamics

Application of the Bernoulli Equation

4.1

In a vertical pipe carrying water, pressure gauges are inserted at points A and B where

the pipe diameters are 0.15m and 0.075m respectively. The point B is 2.5m below A

and when the flow rate down the pipe is 0.02 cumecs, the pressure at B is 14715 N/m2

greater than that at A.

Assuming the losses in the pipe between A and B can be expressed as where v is

the velocity at A, find the value of k.

If the gauges at A and B are replaced by tubes filled with water and connected to a U-

tube containing mercury of relative density 13.6, give a sketch showing how the

levels in the two limbs of the U-tube differ and calculate the value of this difference in

metres.

[k = 0.319, 0.0794m]

4.2

A Venturimeter with an entrance diameter of 0.3m and a throat diameter of 0.2m is

used to measure the volume of gas flowing through a pipe. The discharge coefficient

of the meter is 0.96.

Assuming the specific weight of the gas to be constant at 19.62 N/m3, calculate the

volume flowing when the pressure difference between the entrance and the throat is

measured as 0.06m on a water U-tube manometer.

[0.816 m3/s]

4.3

A Venturimeter is used for measuring flow of water along a pipe. The diameter of the

Venturi throat is two fifths the diameter of the pipe. The inlet and throat are connected

by water filled tubes to a mercury U-tube manometer. The velocity of flow along the

pipe is found to be m/s, where H is the manometer reading in metres of

mercury. Determine the loss of head between inlet and throat of the Venturi when H is

0.49m. (Relative density of mercury is 13.6).

[0.23m of water]

4.4

Water is discharging from a tank through a convergent-divergent mouthpiece. The

exit from the tank is rounded so that losses there may be neglected and the minimum

diameter is 0.05m.

If the head in the tank above the centre-line of the mouthpiece is 1.83m. a) What is

the discharge?

b) What must be the diameter at the exit if the absolute pressure at the minimum area

is to be 2.44m of water? c) What would the discharge be if the divergent part of the

105

mouth piece were removed. (Assume atmospheric pressure is 10m of water).

[0.0752m, 0.0266m3/s, 0.0118m3/s]

4.5

A closed tank has an orifice 0.025m diameter in one of its vertical sides. The tank

contains oil to a depth of 0.61m above the centre of the orifice and the pressure in the

air space above the oil is maintained at 13780 N/m2 above atmospheric. Determine the

discharge from the orifice.

(Coefficient of discharge of the orifice is 0.61, relative density of oil is 0.9).

[0.00195 m3/s]

4.6

The discharge of a Venturimeter was found to be constant for rates of flow exceeding

a certain value. Show that for this condition the loss of head due to friction in the

convergent parts of the meter can be expressed as KQ2 m where K is a constant and Q

is the rate of flow in cumecs.

Obtain the value of K if the inlet and throat diameter of the Venturimeter are 0.102m

and 0.05m respectively and the discharge coefficient is 0.96.

[K=1060]

4.7

A Venturimeter is to fitted in a horizontal pipe of 0.15m diameter to measure a flow

of water which may be anything up to 240m3/hour. The pressure head at the inlet for

this flow is 18m above atmospheric and the pressure head at the throat must not be

lower than 7m below atmospheric. Between the inlet and the throat there is an

estimated frictional loss of 10% of the difference in pressure head between these

points. Calculate the minimum allowable diameter for the throat.

[0.063m]

4.8

A Venturimeter of throat diameter 0.076m is fitted in a 0.152m diameter vertical pipe

in which liquid of relative density 0.8 flows downwards. Pressure gauges are fitted to

the inlet and to the throat sections. The throat being 0.914m below the inlet. Taking

the coefficient of the meter as 0.97 find the discharge

a) when the pressure gauges read the same b)when the inlet gauge reads 15170 N/m2

higher than the throat gauge.

[0.0192m3/s, 0.034m3/s]

Tank emptying

5.1

A reservoir is circular in plan and the sides slope at an angle of tan-1(1/5) to the

horizontal. When the reservoir is full the diameter of the water surface is 50m.

Discharge from the reservoir takes place through a pipe of diameter 0.65m, the outlet

being 4m below top water level. Determine the time for the water level to fall 2m

assuming the discharge to be cumecs where a is the cross sectional area

of the pipe in m2 and H is the head of water above the outlet in m.

[1325 seconds]

106

5.2

A rectangular swimming pool is 1m deep at one end and increases uniformly in depth

to 2.6m at the other end. The pool is 8m wide and 32m long and is emptied through an

orifice of area 0.224m2, at the lowest point in the side of the deep end. Taking Cd for

the orifice as 0.6, find, from first principles,

a) the time for the depth to fall by 1m b) the time to empty the pool completely.

[299 second, 662 seconds]

5.3

A vertical cylindrical tank 2m diameter has, at the bottom, a 0.05m diameter sharp

edged orifice for which the discharge coefficient is 0.6.

a) If water enters the tank at a constant rate of 0.0095 cumecs find the depth of water

above the orifice when the level in the tank becomes stable.

b) Find the time for the level to fall from 3m to 1m above the orifice when the inflow

is turned off.

c) If water now runs into the tank at 0.02 cumecs, the orifice remaining open, find the

rate of rise in water level when the level has reached a depth of 1.7m above the

orifice.

[a) 3.314m, b) 881 seconds, c) 0.252m/min]

5.4

A horizontal boiler shell (i.e. a horizontal cylinder) 2m diameter and 10m long is half

full of water. Find the time of emptying the shell through a short vertical pipe,

diameter 0.08m, attached to the bottom of the shell. Take the coefficient of discharge

to be 0.8.

[1370 seconds]

5.5

Two cylinders standing upright contain liquid and are connected by a submerged

orifice. The diameters of the cylinders are 1.75m and 1.0m and of the orifice, 0.08m.

The difference in levels of the liquid is initially 1.35m. Find how long it will take for

this difference to be reduced to 0.66m if the coefficient of discharge for the orifice is

0.605. (Work from first principles.)

[30.7 seconds]

5.6

A rectangular reservoir with vertical walls has a plan area of 60000m3. Discharge

from the reservoir take place over a rectangular weir. The flow characteristics of the

weir is Q = 0.678 H3/2 cumecs where H is the depth of water above the weir crest. The

sill of the weir is 3.4m above the bottom of the reservoir. Starting with a depth of

water of 4m in the reservoir and no inflow, what will be the depth of water after one

hour? [3.98m]

Notches and weirs

6.1

Deduce an expression for the discharge of water over a right-angled sharp edged V-

notch, given that the coefficient of discharge is 0.61.

A rectangular tank 16m by 6m has the same notch in one of its short vertical sides.

107

Determine the time taken for the head, measured from the bottom of the notch, to fall

from 15cm to 7.5cm.

[1399 seconds]

6.2

Derive an expression for the discharge over a sharp crested rectangular weir. A sharp

edged weir is to be constructed across a stream in which the normal flow is 200

litres/sec. If the maximum flow likely to occur in the stream is 5 times the normal

flow then determine the length of weir necessary to limit the rise in water level to

38.4cm above that for normal flow. Cd=0.61.

[1.24m]

6.3

Show that the rate of flow across a triangular notch is given by Q=CdKH5/2 cumecs,

where Cd is an experimental coefficient, K depends on the angle of the notch, and H is

the height of the undisturbed water level above the bottom of the notch in metres.

State the reasons for the introduction of the coefficient.

Water from a tank having a surface area of 10m2 flows over a 90 notch. It is found

that the time taken to lower the level from 8cm to 7cm above the bottom of the notch

is 43.5seconds. Determine the coefficient Cd assuming that it remains constant during

his period.

[0.635]

6.4

A reservoir with vertical sides has a plan area of 56000m2. Discharge from the

reservoir takes place over a rectangular weir, the flow characteristic of which is

Q=1.77BH3/2 m3/s. At times of maximum rainfall, water flows into the reservoir at the

rate of 9m3/s. Find a) the length of weir required to discharge this quantity if head

must not exceed 0.6m; b) the time necessary for the head to drop from 60cm to 30cm

if the inflow suddenly stops.

[10.94m, 3093seconds]

6.5

Develop a formula for the discharge over a 90 V-notch weir in terms of head above

the bottom of the V.

A channel conveys 300 litres/sec of water. At the outlet end there is a 90 V-notch weir

for which the coefficient of discharge is 0.58. At what distance above the bottom of

the channel should the weir be placed in order to make the depth in the channel

1.30m? With the weir in this position what is the depth of water in the channel when

the flow is 200 litres/sec?

[0.755m, 1.218m]

6.6

Show that the quantity of water flowing across a triangular V-notch of angle 2 is

. Find the flow if the measured head above the bottom of the

V is 38cm, when =45 and Cd=0.6. If the flow is wanted within an accuracy of 2%,

what are the limiting values of the head.

[0.126m3/s, 0.377m, 0.383m]

108

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1. The Momentum Equation And Its Applications

We have all seen moving fluids exerting forces. The lift force on an aircraft is exerted

by the air moving over the wing. A jet of water from a hose exerts a force on

whatever it hits. In fluid mechanics the analysis of motion is performed in the same

way as in solid mechanics - by use of Newton's laws of motion. Account is also taken

for the special properties of fluids when in motion.

The momentum equation is a statement of Newton's Second Law and relates the sum

of the forces acting on an element of fluid to its acceleration or rate of change of

momentum. You will probably recognise the equation F = ma which is used in the

analysis of solid mechanics to relate applied force to acceleration. In fluid mechanics

it is not clear what mass of moving fluid we should use so we use a different form of

the equation.

Newton's 2nd Law can be written:

The Rate of change of momentum of a body is equal to the resultant force acting on

the body, and takes place in the direction of the force.

To determine the rate of change of momentum for a fluid we will consider a

streamtube as we did for the Bernoulli equation,

We start by assuming that we have steady flow which is non-uniform flowing in a

stream tube.

http://www.efm.leeds.ac.uk/CIVE/CIVE1400/course.html


109

A streamtube in three and two-dimensions

In time a volume of the fluid moves from the inlet a distance , so the volume

entering the streamtube in the time is

this has mass,

and momentum

Similarly, at the exit, we can obtain an expression for the momentum leaving the

steamtube:

We can now calculate the force exerted by the fluid using Newton's 2nd Law. The

force is equal to the rate of change of momentum. So

We know from continuity that , and if we have a fluid of constant

density, i.e. , then we can write

110

For an alternative derivation of the same expression, as we know from conservation of

mass in a stream tube that

we can write

The rate at which momentum leaves face 1 is

The rate at which momentum enters face 2 is

Thus the rate at which momentum changes across the stream tube is

i.e.

This force is acting in the direction of the flow of the fluid.

This analysis assumed that the inlet and outlet velocities were in the same direction -

i.e. a one dimensional system. What happens when this is not the case?

Consider the two dimensional system in the figure below:

111

Two dimensional flow in a streamtube

At the inlet the velocity vector, , makes an angle, , with the x-axis, while at the

outlet make an angle . In this case we consider the forces by resolving in the

directions of the co-ordinate axes.

The force in the x-direction

And the force in the y-direction

We then find the resultant force by combining these vectorially:

And the angle which this force acts at is given by

For a three-dimensional (x, y, z) system we then have an extra force to calculate and

resolve in the z-direction. This is considered in exactly the same way.

112

In summary we can say:

Remember that we are working with vectors so F is in the direction of the velocity.

This force is made up of three components:

Force exerted on the fluid by any solid body touching the control volume

Force exerted on the fluid body (e.g. gravity)

Force exerted on the fluid by fluid pressure outside the control volume

So we say that the total force, FT, is given by the sum of these forces:

The force exerted by the fluid on the solid body touching the control volume is

opposite to . So the reaction force, R, is given by



Application of the Momentum Equation

In this section we will consider the following examples:

1. Force due to the flow of fluid round a pipe bend.

2. Force on a nozzle at the outlet of a pipe.

3. Impact of a jet on a plane surface.

1. The force due the flow around a pipe bend

Consider a pipe bend with a constant cross section lying in the horizontal plane and

turning through an angle of .



113

Flow round a pipe bend of constant cross-section

Why do we want to know the forces here? Because the fluid changes direction, a force

(very large in the case of water supply pipes,) will act in the bend. If the bend is not

fixed it will move and eventually break at the joints. We need to know how much

force a support (thrust block) must withstand.

Step in Analysis:

1. Draw a control volume

2. Decide on co-ordinate axis system

3. Calculate the total force

4. Calculate the pressure force

5. Calculate the body force

6. Calculate the resultant force

1 Control Volume

The control volume is draw in the above figure, with faces at the inlet and outlet of

the bend and encompassing the pipe walls.

2 Co-ordinate axis system

114

It is convenient to choose the co-ordinate axis so that one is pointing in the direction

of the inlet velocity. In the above figure the x-axis points in the direction of the inlet

velocity.

3 Calculate the total force

In the x-direction:

In the y-direction:

4 Calculate the pressure force

5 Calculate the body force

There are no body forces in the x or y directions. The only body force is that exerted

by gravity (which acts into the paper in this example - a direction we do not need to

consider).

6 Calculate the resultant force

And the resultant force on the fluid is given by

115

And the direction of application is

the force on the bend is the same magnitude but in the opposite direction

2. Force on a pipe nozzle

Force on the nozzle at the outlet of a pipe. Because the fluid is contracted at the

nozzle forces are induced in the nozzle. Anything holding the nozzle (e.g. a fireman)

must be strong enough to withstand these forces.

The analysis takes the same procedure as above:







1 & 2 Control volume and Co-ordinate axis are shown in the figure below.

116

Notice how this is a one dimensional system which greatly simplifies matters.


By continuity, , so

4 Calculate the pressure force

We use the Bernoulli equation to calculate the pressure

Is friction losses are neglected,

the nozzle is horizontal,

and the pressure outside is atmospheric, ,

and with continuity gives


117

The only body force is the weight due to gravity in the y-direction - but we need not

consider this as the only forces we are considering are in the x-direction.


So the fireman must be able to resist the force of

3. Impact of a Jet on a Plane

We will first consider a jet hitting a flat plate (a plane) at an angle of 90, as shown in

the figure below.

We want to find the reaction force of the plate i.e. the force the plate will have to

apply to stay in the same position.

A perpendicular jet hitting a plane.

The analysis take the same procedure as above:







118



As the system is symmetrical the forces in the y-direction cancel i.e.

4 Calculate the pressure force.

The pressure force is zero as the pressure at both the inlet and the outlets to the

control volume are atmospheric.


As the control volume is small we can ignore the body force due to the weight of

gravity.


Exerted on the fluid.

The force on the plane is the same magnitude but in the opposite direction

119



More Applications of the Momentum Equation

In this section we will consider the following examples:

1. Force due to flow round a curved vane.

2. A curved vane on a Pelton wheel turbine.

3. Impact of a jet on An angled plane surface.

1. Force on a curved vane

This case is similar to that of a pipe, but the analysis is simpler because the pressures

are equal - atmospheric , and both the cross-section and velocities (in the direction of

flow) remain constant. The jet, vane and co-ordinate direction are arranged as in the

figure below.

Jet deflected by a curved vane.

1 & 2 Control volume and Co-ordinate axis are shown in the figure above.

3 Calculate the total force in the x direction



120

but , so

and in the y-direction


Again, the pressure force is zero as the pressure at both the inlet and the outlets to the



No body forces in the x-direction, = 0.

In the y-direction the body force acting is the weight of the fluid. If V is the volume of

the fluid on he vane then,

(This is often small is the jet volume is small and sometimes ignored in analysis.)




121

exerted on the fluid.

The force on the vane is the same magnitude but in the opposite direction

2. Pelton wheel blade

The above analysis of impact of jets on vanes can be extended and applied to analysis

of turbine blades. One particularly clear demonstration of this is with the blade of a

turbine called the pelton wheel. The arrangement of a pelton wheel is shown in the

figure below. A narrow jet (usually of water) is fired at blades which stick out around

the periphery of a large metal disk. The shape of each of these blade is such that as the

jet hits the blade it splits in two (see figure below) with half the water diverted to one

side and the other to the other. This splitting of the jet is beneficial to the turbine

mounting - it causes equal and opposite forces (hence a sum of zero) on the bearings.

Pelton wheel arrangement and jet hitting cross-section of blade.

A closer view of the blade and control volume used for analysis can be seen in the

figure below.

Analysis again take the following steps:








122

3 Calculate the total force in the x direction

and in the y-direction it is symmetrical, so


The pressure force is zero as the pressure at both the inlet and the outlets to the



We are only considering the horizontal plane in which there are no body forces.



The force on the blade is the same magnitude but in the opposite direction

123

So the blade moved in the x-direction.

In a real situation the blade is moving. The analysis can be extended to include this by

including the amount of momentum entering the control volume over the time the

blade remains there. This will be covered in the level 2 module next year.

3. Force due to a jet hitting an inclined plane

We have seen above the forces involved when a jet hits a plane at right angles. If the

plane is tilted to an angle the analysis becomes a little more involved. This is

demonstrated below.

A jet hitting an inclined plane.

(Note that for simplicity gravity and friction will be neglected from this analysis.)

We want to find the reaction force normal to the plate so we choose the axis system as

above so that is normal to the plane. The diagram may be rotated to align it with these

axes and help comprehension, as shown below

124

Rotated view of the jet hitting the inclined plane.

We do not know the velocities of flow in each direction. To find these we can apply

Bernoulli equation

The height differences are negligible i.e. z1 = z2 = z3 and the pressures are all

atmospheric = 0. So

. u1 = u2 = u3 = u

By continuity

Q1 = Q2 + Q3

u1A1 = u2A2 + u3A3

so

A1 = A2 + A3

Q1 = A1u

Q2 = A2u

Q3 = (A1 - A2)u

Using this we can calculate the forces in the same way as before.





1 Calculate the total force in the x-direction.

125

Remember that the co-ordinate system is normal to the plate.

but as the jets are parallel to the plate with no component in the x-

direction.

, so


All zero as the pressure is everywhere atmospheric.


As the control volume is small, hence the weight of fluid is small, we can ignore the

body forces.



The force on the plate is the same magnitude but in the opposite direction

We can find out how much discharge goes along in each direction on the plate. Along

the plate, in the y-direction, the total force must be zero, .

Also in the y-direction: , so

As forces parallel to the plate are zero,

126

From above

and from above we have so

as u2 = u3 = u

So we know the discharge in each direction



Dynamics


6.1

The figure below shows a smooth curved vane attached to a rigid foundation. The jet

of water, rectangular in section, 75mm wide and 25mm thick, strike the vane with a

velocity of 25m/s. Calculate the vertical and horizontal components of the force

exerted on the vane and indicate in which direction these components act.

[Horizontal 233.4 N acting from right to left. Vertical 1324.6 N acting downwards]



127

6.2

A 600mm diameter pipeline carries water under a head of 30m with a velocity of

3m/s. This water main is fitted with a horizontal bend which turns the axis of the

pipeline through 75 (i.e. the internal angle at the bend is 105). Calculate the resultant

force on the bend and its angle to the horizontal.

[104.044 kN, 52 29']

6.3

A horizontal jet of water 2103 mm2 cross-section and flowing at a velocity of 15 m/s

hits a flat plate at 60 to the axis (of the jet) and to the horizontal. The jet is such that

there is no side spread. If the plate is stationary, calculate a) the force exerted on the

plate in the direction of the jet and b) the ratio between the quantity of fluid that is

deflected upwards and that downwards. (Assume that there is no friction and therefore

no shear force.)

[338N, 3:1]

6.4

A 75mm diameter jet of water having a velocity of 25m/s strikes a flat plate, the

normal of which is inclined at 30 to the jet. Find the force normal to the surface of the

plate.

[2.39kN]

6.5

The outlet pipe from a pump is a bend of 45 rising in the vertical plane (i.e. and

internal angle of 135). The bend is 150mm diameter at its inlet and 300mm diameter

at its outlet. The pipe axis at the inlet is horizontal and at the outlet it is 1m higher. By

neglecting friction, calculate the force and its direction if the inlet pressure is

100kN/m2 and the flow of water through the pipe is 0.3m3/s. The volume of the pipe

is 0.075m3.

[13.94kN at 67 40' to the horizontal]

6.6

The force exerted by a 25mm diameter jet against a flat plate normal to the axis of the

jet is 650N. What is the flow in m3/s?

[0.018 m3/s]

6.7

A curved plate deflects a 75mm diameter jet through an angle of 45. For a velocity in

the jet of 40m/s to the right, compute the components of the force developed against

the curved plate. (Assume no friction).

[Rx=2070N, Ry=5000N down]

128

6.8

A 45 reducing bend, 0.6m diameter upstream, 0.3m diameter downstream, has water

flowing through it at the rate of 0.45m3/s under a pressure of 1.45 bar. Neglecting any

loss is head for friction, calculate the force exerted by the water on the bend, and its

direction of application.

[R=34400N to the right and down, = 14]



1. Real fluids

The flow of real fluids exhibits viscous effect, that is they tend to "stick" to solid

surfaces and have stresses within their body.

You might remember from earlier in the course Newtons law of viscosity:

This tells us that the shear stress, , in a fluid is proportional to the velocity gradient -

the rate of change of velocity across the fluid path. For a "Newtonian" fluid we can

write:

where the constant of proportionality, is known as the coefficient of viscosity (or

simply viscosity). We saw that for some fluids - sometimes known as exotic fluids -

the value of changes with stress or velocity gradient. We shall only deal with

Newtonian fluids.

In his lecture we shall look at how the forces due to momentum changes on the fluid

and viscous forces compare and what changes take place.

2. Laminar and turbulent flow

If we were to take a pipe of free flowing water and inject a dye into the middle of the

stream, what would we expect to happen?

This



129

this

or this

Actually both would happen - but for different flow rates. The top occurs when the

fluid is flowing fast and the lower when it is flowing slowly.

The top situation is known as turbulent flow and the lower as laminar flow.

In laminar flow the motion of the particles of fluid is very orderly with all particles

moving in straight lines parallel to the pipe walls.

130

But what is fast or slow? And at what speed does the flow pattern change? And why

might we want to know this?

The phenomenon was first investigated in the 1880s by Osbourne Reynolds in an

experiment which has become a classic in fluid mechanics.

He used a tank arranged as above with a pipe taking water from the centre into which

he injected a dye through a needle. After many experiments he saw that this

expression

where = density, u = mean velocity, d = diameter and = viscosity

would help predict the change in flow type. If the value is less than about 2000 then

flow is laminar, if greater than 4000 then turbulent and in between these then in the

transition zone.

This value is known as the Reynolds number, Re:

131

Laminar flow: Re < 2000

Transitional flow: 2000 < Re < 4000

Turbulent flow: Re > 4000

What are the units of this Reynolds number? We can fill in the equation with SI units:

i.e. it has no units. A quantity that has no units is known as a non-dimensional (or

dimensionless) quantity. Thus the Reynolds number, Re, is a non-dimensional

number.

We can go through an example to discover at what velocity the flow in a pipe stops

being laminar.

If the pipe and the fluid have the following properties:

water density = 1000 kg/m3

pipe diameter d = 0.5m

(dynamic) viscosity, = 0.55x103 Ns/m2

We want to know the maximum velocity when the Re is 2000.

If this were a pipe in a house central heating system, where the pipe diameter is

typically 0.015m, the limiting velocity for laminar flow would be, 0.0733 m/s.

Both of these are very slow. In practice it very rarely occurs in a piped water system -

the velocities of flow are much greater. Laminar flow does occur in situations with

fluids of greater viscosity - e.g. in bearing with oil as the lubricant.

132

At small values of Re above 2000 the flow exhibits small instabilities. At values of

about 4000 we can say that the flow is truly turbulent. Over the past 100 years since

this experiment, numerous more experiments have shown this phenomenon of limits

of Re for many different Newtonian fluids - including gasses.

What does this abstract number mean?

We can say that the number has a physical meaning, by doing so it helps to

understand some of the reasons for the changes from laminar to turbulent flow.

It can be interpreted that when the inertial forces dominate over the viscous forces

(when the fluid is flowing faster and Re is larger) then the flow is turbulent. When the

viscous forces are dominant (slow flow, low Re) they are sufficient enough to keep all

the fluid particles in line, then the flow is laminar.

In summary:

Laminar flow

Re < 2000

'low' velocity

Dye does not mix with water

Fluid particles move in straight lines

Simple mathematical analysis possible

Rare in practice in water systems.

Transitional flow

2000 > Re < 4000

'medium' velocity

Dye stream wavers in water - mixes slightly.

Turbulent flow

Re > 4000

'high' velocity

Dye mixes rapidly and completely

Particle paths completely irregular

Average motion is in the direction of the flow

Cannot be seen by the naked eye

Changes/fluctuations are very difficult to detect. Must use laser.

Mathematical analysis very difficult - so experimental measures are used

Most common type of flow.

133

3. Pressure loss due to friction in a pipeline.

Up to this point on the course we have considered ideal fluids where there have been

no losses due to friction or any other factors. In reality, because fluids are viscous,

energy is lost by flowing fluids due to friction which must be taken into account. The

effect of the friction shows itself as a pressure (or head) loss.

In a pipe with a real fluid flowing, at the wall there is a shearing stress retarding the

flow, as shown below.

If a manometer is attached as the pressure (head) difference due to the energy lost by

the fluid overcoming the shear stress can be easily seen.

The pressure at 1 (upstream) is higher than the pressure at 2.

We can do some analysis to express this loss in pressure in terms of the forces acting

on the fluid.

Consider a cylindrical element of incompressible fluid flowing in the pipe, as shown

134

The pressure at the upstream end is p, and at the downstream end the pressure has

fallen by p to (p-p).

The driving force due to pressure (F = Pressure x Area) can then be written

driving force = Pressure force at 1 - pressure force at 2

The retarding force is that due to the shear stress by the walls

As the flow is in equilibrium,

driving force = retarding force

Giving an expression for pressure loss in a pipe in terms of the pipe diameter and the

shear stress at the wall on the pipe.

135

The shear stress will vary with velocity of flow and hence with Re. Many experiments

have been done with various fluids measuring the pressure loss at various Reynolds

numbers. These results plotted to show a graph of the relationship between pressure

loss and Re look similar to the figure below:

This graph shows that the relationship between pressure loss and Re can be expressed

as

As these are empirical relationships, they help in determining the pressure loss but not

in finding the magnitude of the shear stress at the wall w on a particular fluid. If we

knew w we could then use it to give a general equation to predict the pressure loss.

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4. Pressure loss during laminar flow in a pipe

In general the shear stress w. is almost impossible to measure. But for laminar flow it

is possible to calculate a theoretical value for a given velocity, fluid and pipe

dimension.

In laminar flow the paths of individual particles of fluid do not cross, so the flow may

be considered as a series of concentric cylinders sliding over each other - rather like

the cylinders of a collapsible pocket telescope.

As before, consider a cylinder of fluid, length L, radius r, flowing steadily in the

centre of a pipe.

We are in equilibrium, so the shearing forces on the cylinder equal the pressure

forces.

By Newtons law of viscosity we have , where y is the distance from the

wall. As we are measuring from the pipe centre then we change the sign and replace y

with r distance from the centre, giving

Which can be combined with the equation above to give

In an integral form this gives an expression for velocity,

137

Integrating gives the value of velocity at a point distance r from the centre

At r = 0, (the centre of the pipe), u = umax, at r = R (the pipe wall) u = 0, giving

so, an expression for velocity at a point r from the pipe centre when the flow is

laminar is

Note how this is a parabolic profile (of the form y = ax2 + b ) so the velocity profile in

the pipe looks similar to the figure below

What is the discharge in the pipe?

138

So the discharge can be written

This is the Hagen-Poiseuille equation for laminar flow in a pipe. It expresses the

discharge Q in terms of the pressure gradient ( ), diameter of the pipe and the

viscosity of the fluid.

We are interested in the pressure loss (head loss) and want to relate this to the velocity

of the flow. Writing pressure loss in terms of head loss hf, i.e. p = ghf

This shows that pressure loss is directly proportional to the velocity when flow is

laminar.

It has been validated many time by experiment.

It justifies two assumptions:

1. fluid does not slip past a solid boundary

2. Newtons hypothesis.



1. Boundary Layers

(Recommended extra reading for this section: Fluid Mechanics by Douglas J F,

Gasiorek J M, and Swaffield J A. Longman publishers. Pages 327-332.)

When a fluid flows over a stationary surface, e.g. the bed of a river, or the wall of a

pipe, the fluid touching the surface is brought to rest by the shear stress o at the wall.

The velocity increases from the wall to a maximum in the main stream of the flow.



139

Looking at this two-dimensionally we get the above velocity profile from the wall to

the centre of the flow.

This profile doesn't just exit, it must build up gradually from the point where the fluid

starts to flow past the surface - e.g. when it enters a pipe.

If we consider a flat plate in the middle of a fluid, we will look at the build up of the

velocity profile as the fluid moves over the plate.

Upstream the velocity profile is uniform, (free stream flow) a long way downstream

we have the velocity profile we have talked about above. This is the known as fully

developed flow. But how do we get to that state?

This region, where there is a velocity profile in the flow due to the shear stress at the

wall, we call the boundary layer. The stages of the formation of the boundary layer

are shown in the figure below:

140

We define the thickness of this boundary layer as the distance from the wall to the

point where the velocity is 99% of the "free stream" velocity, the velocity in the

middle of the pipe or river.

boundary layer thickness, = distance from wall to point where u = 0.99 umainstream

The value of will increase with distance from the point where the fluid first starts to

pass over the boundary - the flat plate in our example. It increases to a maximum in

fully developed flow.

Correspondingly, the drag force D on the fluid due to shear stress o at the wall

increases from zero at the start of the plate to a maximum in the fully developed flow

region where it remains constant. We can calculate the magnitude of the drag force by

using the momentum equation. But this complex and not necessary for this course.

Our interest in the boundary layer is that its presence greatly affects the flow through

or round an object. So here we will examine some of the phenomena associated with

the boundary layer and discuss why these occur.

2. Formation of the boundary layer

Above we noted that the boundary layer grows from zero when a fluid starts to flow

over a solid surface. As is passes over a greater length more fluid is slowed by friction

between the fluid layers close to the boundary. Hence the thickness of the slower

layer increases.

The fluid near the top of the boundary layer is dragging the fluid nearer to the solid

surface along. The mechanism for this dragging may be one of two types:

The first type occurs when the normal viscous forces (the forces which hold the fluid

together) are large enough to exert drag effects on the slower moving fluid close to

the solid boundary. If the boundary layer is thin then the velocity gradient normal to

the surface, (du/dy), is large so by Newton's law of viscosity the shear stress, =

(du/dy), is also large. The corresponding force may then be large enough to exert drag

on the fluid close to the surface.

As the boundary layer thickness becomes greater, so the velocity gradient become

smaller and the shear stress decreases until it is no longer enough to drag the slow

fluid near the surface along. If this viscous force was the only action then the fluid

would come to a rest.

It, of course, does not come to rest but the second mechanism comes into play. Up to

this point the flow has been laminar and Newton's law of viscosity has applied. This

part of the boundary layer is known as the laminar boundary layer

The viscous shear stresses have held the fluid particles in a constant motion within

layers. They become small as the boundary layer increases in thickness and the

velocity gradient gets smaller. Eventually they are no longer able to hold the flow in

layers and the fluid starts to rotate.

141

This causes the fluid motion to rapidly becomes turbulent. Fluid from the fast moving

region moves to the slower zone transferring momentum and thus maintaining the

fluid by the wall in motion. Conversely, slow moving fluid moves to the faster

moving region slowing it down. The net effect is an increase in momentum in the

boundary layer. We call the part of the boundary layer the turbulent boundary layer.

At points very close to the boundary the velocity gradients become very large and the

velocity gradients become very large with the viscous shear forces again becoming

large enough to maintain the fluid in laminar motion. This region is known as the

laminar sub-layer. This layer occurs within the turbulent zone and is next to the wall

and very thin - a few hundredths of a mm.

3. Surface roughness effect

Despite its thinness, the laminar sub-layer can play a vital role in the friction

characteristics of the surface.

This is particularly relevant when defining pipe friction - as will be seen in more

detail in the level 2 module. In turbulent flow if the height of the roughness of a pipe

is greater than the thickness of the laminar sub-layer then this increases the amount of

turbulence and energy losses in the flow. If the height of roughness is less than the

thickness of the laminar sub-layer the pipe is said to be smooth and it has little effect

on the boundary layer.

In laminar flow the height of roughness has very little effect

4. Boundary layers in pipes

As flow enters a pipe the boundary layer will initially be of the laminar form. This

will change depending on the ration of inertial and viscous forces; i.e. whether we

have laminar (viscous forces high) or turbulent flow (inertial forces high).

From earlier we saw how we could calculate whether a particular flow in a pipe is

laminar or turbulent using the Reynolds number.

142

= density u = velocity = viscosity d = pipe diameter)

Laminar flow: Re < 2000

Transitional flow: 2000 < Re < 4000

Turbulent flow: Re > 4000

If we only have laminar flow the profile is parabolic - as proved in earlier lectures - as

only the first part of the boundary layer growth diagram is used. So we get the top

diagram in the above figure.

If turbulent (or transitional), both the laminar and the turbulent (transitional) zones of

the boundary layer growth diagram are used. The growth of the velocity profile is thus

like the bottom diagram in the above figure.

Once the boundary layer has reached the centre of the pipe the flow is said to be fully

developed. (Note that at this point the whole of the fluid is now affected by the

boundary friction.)

The length of pipe before fully developed flow is achieved is different for the two

types of flow. The length is known as the entry length.

Laminar flow entry length 120 diameter

Turbulent flow entry length 60 diameter

143

5. Boundary layer separation

Convergent flows: Negative pressure gradients

If flow over a boundary occurs when there is a pressure decrease in the direction of

flow, the fluid will accelerate and the boundary layer will become thinner.

This is the case for convergent flows.

The accelerating fluid maintains the fluid close to the wall in motion. Hence the flow

remains stable and turbulence reduces. Boundary layer separation does not occur.

Divergent flows: Positive pressure gradients

When the pressure increases in the direction of flow the situation is very different.

Fluid outside the boundary layer has enough momentum to overcome this pressure

which is trying to push it backwards. The fluid within the boundary layer has so little

momentum that it will very quickly be brought to rest, and possibly reversed in

direction. If this reversal occurs it lifts the boundary layer away from the surface as

shown below.

144

This phenomenon is known as boundary layer separation.

At the edge of the separated boundary layer, where the velocities change direction, a

line of vortices occur (known as a vortex sheet). This happens because fluid to either

side is moving in the opposite direction.

This boundary layer separation and increase in the turbulence because of the vortices

results in very large energy losses in the flow.

These separating / divergent flows are inherently unstable and far more energy is lost

than in parallel or convergent flow.

6. Examples of boundary layer separation

A divergent duct or diffuser

The increasing area of flow causes a velocity drop (according to continuity) and hence

a pressure rise (according to the Bernoulli equation).

145

Increasing the angle of the diffuser increases the probability of boundary layer

separation. In a Venturi meter it has been found that an angle of about 6 provides the

optimum balance between length of meter and danger of boundary layer separation

which would cause unacceptable pressure energy losses.

Tee-Junctions

Assuming equal sized pipes, as fluid is removed, the velocities at 2 and 3 are smaller

than at 1, the entrance to the tee. Thus the pressure at 2 and 3 are higher than at 1.

These two adverse pressure gradients can cause the two separations shown in the

diagram above.

Y-Junctions

Tee junctions are special cases of the Y-junction with similar separation zones

occurring. See the diagram below.

146

Downstream, away from the junction, the boundary layer reattaches and normal flow

occurs i.e. the effect of the boundary layer separation is only local. Nevertheless fluid

downstream of the junction will have lost energy.

Bends

Two separation zones occur in bends as shown above. The pressure at b must be

greater than at a as it must provide the required radial acceleration for the fluid to get

round the bend. There is thus an adverse pressure gradient between a and b so

separation may occur here.

Pressure at c is less than at the entrance to the bend but pressure at d has returned to

near the entrance value - again this adverse pressure gradient may cause boundary

layer separation.

Flow past a cylinder

The pattern of flow around a cylinder varies with the velocity of flow. If flow is very

slow with the Reynolds number ( v diameter/ less than 0.5, then there is no

separation of the boundary layers as the pressure difference around the cylinder is

very small. The pattern is something like that in the figure below.

147

If 2 < Re < 70 then the boundary layers separate symmetrically on either side of the

cylinder. The ends of these separated zones remain attached to the cylinder, as shown

below.

Above a Re of 70 the ends of the separated zones curl up into vortices and detach

alternately from each side forming a trail of vortices on the down stream side of the

cylinder. This trial in known as a Karman vortex trail or street. This vortex trail can

easily be seen in a river by looking over a bridge where there is a pier to see the line

of vortices flowing away from the bridge. The phenomenon is responsible for the

whistling of hanging telephone or power cables. A more significant event was the

famous failure of the Tacoma narrows bridge. Here the frequency of the alternate

vortex shedding matched the natural frequency of the bridge deck and resonance

amplified the vibrations until the bridge collapsed. (The frequency of vortex shedding

from a cylinder can be predicted. We will not try to predict it here but a derivation of

the expression can be found in many fluid mechanics text books.)

148

Looking at the figure above, the formation of the separation occurs as the fluid

accelerates from the centre to get round the cylinder (it must accelerate as it has

further to go than the surrounding fluid). It reaches a maximum at Y, where it also has

also dropped in pressure. The adverse pressure gradient between here and the

downstream side of the cylinder will cause the boundary layer separation if the flow is

fast enough, (Re > 2.)

Aerofoil

Normal flow over a aerofoil (a wing cross-section) is shown in the figure below with

the boundary layers greatly exaggerated.

The velocity increases as air it flows over the wing. The pressure distribution is

similar to that shown below so transverse lift force occurs.

149

If the angle of the wing becomes too great and boundary layer separation occurs on

the top of the aerofoil the pressure pattern will change dramatically. This phenomenon

is known as stalling.

When stalling occurs, all, or most, of the 'suction' pressure is lost, and the plane will

suddenly drop from the sky! The only solution to this is to put the plane into a dive to

regain the boundary layer. A transverse lift force is then exerted on the wing which

gives the pilot some control and allows the plane to be pulled out of the dive.

Fortunately there are some mechanisms for preventing stalling. They all rely on

preventing the boundary layer from separating in the first place.

1. Arranging the engine intakes so that they draw slow air from the boundary

layer at the rear of the wing though small holes helps to keep the boundary

layer close to the wing. Greater pressure gradients can be maintained before

separation take place.

2. Slower moving air on the upper surface can be increased in speed by bringing

air from the high pressure area on the bottom of the wing through slots.

Pressure will decrease on the top so the adverse pressure gradient which would

cause the boundary layer separation reduces.

150

3. Putting a flap on the end of the wing and tilting it before separation occurs

increases the velocity over the top of the wing, again reducing the pressure and

chance of separation occurring.



1. Dimensional Analysis

In engineering the application of fluid mechanics in designs make much of the use of

empirical results from a lot of experiments. This data is often difficult to present in a

readable form. Even from graphs it may be difficult to interpret. Dimensional analysis

provides a strategy for choosing relevant data and how it should be presented.

This is a useful technique in all experimentally based areas of engineering. If it is

possible to identify the factors involved in a physical situation, dimensional analysis

can form a relationship between them.

The resulting expressions may not at first sight appear rigorous but these qualitative

results converted to quantitative forms can be used to obtain any unknown factors

from experimental analysis.

2. Dimensions and units

Any physical situation can be described by certain familiar properties e.g. length,

velocity, area, volume, acceleration etc. These are all known as dimensions.

Of course dimensions are of no use without a magnitude being attached. We must

know more than that something has a length. It must also have a standardised unit -

such as a meter, a foot, a yard etc.

Dimensions are properties which can be measured. Units are the standard elements we

use to quantify these dimensions.



151

In dimensional analysis we are only concerned with the nature of the dimension i.e. its

quality not its quantity. The following common abbreviation are used:

length = L

mass = M

time = T

force = F

temperature =

In this module we are only concerned with L, M, T and F (not ). We can represent

all the physical properties we are interested in with L, T and one of M or F (F can be

represented by a combination of LTM). These notes will always use the LTM

combination.

The following table (taken from earlier in the course) lists dimensions of some

common physical quantities:

Quantity SI Unit . Dimension

velocity m/s ms-1 LT-1

acceleration m/s2 ms-2 LT-2

force N

kg m/s2

kg ms-2

M LT-2

energy (or work)

Joule J

N m,

kg m2/s2

kg m2s-2

ML2T-2

power

Watt W

N m/s

kg m2/s3

Nms-1

kg m2s-3

ML2T-3

pressure ( or stress)

Pascal P,

N/m2,

kg/m/s2

Nm-2

kg m-1s-2

ML-1T-2

density kg/m3 kg m-3 ML-3

specific weight N/m3

kg/m2/s2

kg m-2s-2

ML-2T-2

relative density a ratio

no units .

1

no dimension

viscosity N s/m2

kg/m s

N sm-2

kg m-1s-1

M L-1T-1

surface tension N/m

kg /s2

Nm-1

kg s-2

MT-2

1. 3. Dimensional Homogeneity

152

Any equation describing a physical situation will only be true if both sides have the

same dimensions. That is it must be dimensionally homogenous.

For example the equation which gives for over a rectangular weir (derived earlier in

this module) is,

The SI units of the left hand side are m3s-1. The units of the right hand side must be

the same. Writing the equation with only the SI units gives

i.e. the units are consistent.

To be more strict, it is the dimensions which must be consistent (any set of units can

be used and simply converted using a constant). Writing the equation again in terms

of dimensions,

Notice how the powers of the individual dimensions are equal, (for L they are both 3,

for T both -1).

This property of dimensional homogeneity can be useful for:

1. Checking units of equations;

2. Converting between two sets of units;

3. Defining dimensionless relationships (see below).

4. Results of dimensional analysis

The result of performing dimensional analysis on a physical problem is a single

equation. This equation relates all of the physical factors involved to one another.

This is probably best seen in an example.

If we want to find the force on a propeller blade we must first decide what might

influence this force.

It would be reasonable to assume that the force, F, depends on the following physical

properties:

diameter, d

forward velocity of the propeller (velocity of the plane), u

153

fluid density,

revolutions per second, N

fluid viscosity,

Before we do any analysis we can write this equation:

F = ( d, u, , N, )

or

0 = ( F, d, u, , N, )

where and 1 are unknown functions.

These can be expanded into an infinite series which can itself be reduced to

F = dm up Nr

where K is some constant and m, p, q, r, s are unknown constant powers.

From dimensional analysis we

1. obtain these powers

2. form the variables into several dimensionless groups

The value of K or the functions and 1 must be determined from experiment. The

knowledge of the dimensionless groups often helps in deciding what experimental

measurements should be taken.

5. Buckingham's theorems

Although there are other methods of performing dimensional analysis, (notably the

indicial method) the method based on the Buckingham theorems gives a good

generalised strategy for obtaining a solution. This will be outlined below.

There are two theorems accredited to Buckingham, and know as his theorems.

1st theorem:

A relationship between m variables (physical properties such as velocity, density etc.)

can be expressed as a relationship between m-n non-dimensional groups of variables

(called groups), where n is the number of fundamental dimensions (such as mass,

length and time) required to express the variables.

154

So if a physical problem can be expressed:

( Q1 , Q2 , Q3 ,………, Qm ) = 0

then, according to the above theorem, this can also be expressed

( 1 , 2 , 3 ,………, Qm-n ) = 0

In fluids, we can normally take n = 3 (corresponding to M, L, T).

2nd theorem

Each group is a function of n governing or repeating variables plus one of the

remaining variables.

6. Choice of repeating variables

Repeating variables are those which we think will appear in all or most of the

groups, and are a influence in the problem. Before commencing analysis of a problem

one must choose the repeating variables. There is considerable freedom allowed in the

choice.

Some rules which should be followed are

1. From the 2nd theorem there can be n ( = 3) repeating variables.

2. When combined, these repeating variables variable must contain all of

dimensions (M, L, T)

(That is not to say that each must contain M,L and T).

3. A combination of the repeating variables must not form a dimensionless

group.

4. The repeating variables do not have to appear in all groups.

5. The repeating variables should be chosen to be measurable in an experimental

investigation. They should be of major interest to the designer. For example,

pipe diameter (dimension L) is more useful and measurable than roughness

height (also dimension L).

In fluids it is usually possible to take , u and d as the threee repeating variables.

This freedom of choice results in there being many different groups which can be

formed - and all are valid. There is not really a wrong choice.

7. An example

Taking the example discussed above of force F induced on a propeller blade, we have

the equation

155

0 = ( F, d, u, , N, )

n = 3 and m = 6

There are m - n = 3 groups, so

( 1 , 2 , 3 ) = 0

The choice of , u, d as the repeating variables satisfies the criteria above. They are

measurable, good design parameters and, in combination, contain all the dimension

M,L and T. We can now form the three groups according to the 2nd theorem,

As the groups are all dimensionless i.e. they have dimensions M0L0T0 we can use

the principle of dimensional homogeneity to equate the dimensions for each group.

For the first group,

In terms of SI units

And in terms of dimensions

For each dimension (M, L or T) the powers must be equal on both sides of the

equation, so

for M: 0 = a1 + 1

a1 = -1

for L: 0 = -3a1 + b1 + c1 + 1

0 = 4 + b1 + c1

for T: 0 = -b1 - 2

b1 = -2

c1 = -4 - b1 = -2

Giving 1 as

156

And a similar procedure is followed for the other groups. Group


equation, so

for M: 0 = a2

for L: 0 = -3a2 + b2 + c2

0 = b2 + c2

for T: 0 = -b2 - 1

b2 = -1

c2 = 1

Giving 2 as

And for the third,


equation, so

for M: 0 = a3 + 1

a3 = -1

for L: 0 = -3a3 + b3 + c3 -1

b3 + c3 = -2

for T: 0 = -b3 - 1

157

b3 = -1

c3 = -1

Giving 3 as

Thus the problem may be described by the following function of the three non-

dimensional groups,

( 1 , 2 , 3 ) = 0

This may also be written:

8. Wrong choice of physical properties.

If, when defining the problem, extra - unimportant - variables are introduced then

extra groups will be formed. They will play very little role influencing the physical

behaviour of the problem concerned and should be identified during experimental

work. If an important / influential variable was missed then a group would be

missing. Experimental analysis based on these results may miss significant

behavioural changes. It is therefore, very important that the initial choice of variables

is carried out with great care.

9. Manipulation of the groups

Once identified manipulation of the groups is permitted. These manipulations do

not change the number of groups involved, but may change their appearance

drastically.

Taking the defining equation as: ( 1 , 2 , 3 ……… m-n ) = 0

Then the following manipulations are permitted:

1. Any number of groups can be combined by multiplication or division to form

a new group which replaces one of the existing. E.g. 1 and 2 may be

combined to form 1a = 1 / 2 so the defining equation becomes

( 1a , 2 , 3 ……… m-n ) = 0

158

2. The reciprocal of any dimensionless group is valid. So ( 1 ,1/ 2 , 3 ………

1/m-n ) = 0 is valid.

3. Any dimensionless group may be raised to any power. So ( (1 )2, (2 )

1/2,

(3 )3……… m-n ) = 0 is valid.

4. Any dimensionless group may be multiplied by a constant.

5. Any group may be expressed as a function of the other groups, e.g.

2 = ( 1 , 3 ……… m-n )

In general the defining equation could look like

( 1 , 1/2 ,( 3 )i……… 0.5m-n ) = 0

10. Common groups

During dimensional analysis several groups will appear again and again for different

problems. These often have names. You will recognise the Reynolds number ud/.

Some common non-dimensional numbers (groups) are listed below.

Reynolds number inertial, viscous force ratio

Euler number pressure, inertial force ratio

Froude number inertial, gravitational force ratio

Weber number inertial, surface tension force ratio

Mach number Local velocity, local velocity of sound ratio

11. Examples

The discharge Q through an orifice is a function of the diameter d, the pressure

difference p, the density , and the viscosity , show that ,

where is some unknown function.

Write out the dimensions of the variables

ML-3 u: LT-1

d: L ML-1T-1

159

p:(force/area) ML-1T-2

We are told from the question that there are 5 variables involved in the problem: d, p,

, and Q.

Choose the three recurring (governing) variables; Q, d,

From Buckingham's theorem we have m-n = 5 - 3 = 2 non-dimensional groups.

For the first group, 1:

M] 0 = c1 + 1

c1 = -1

L] 0 = 3a1 + b1 - 3c1 - 1

-2 = 3a1 + b1

T] 0 = -a1 - 1

a1 = -1

b1 = 1

And the second group 2 :

(note p is a pressure (force/area) with dimensions ML-1T-2)

M] 0 = c2 + 1

c2 = -1

160

L] 0 = 3a2 + b2 - 3c2 - 1

-2 = 3a2 + b2

T] 0 = -a2 - 2

a2 = - 2

b2 = 4

So the physical situation is described by this function of non-dimensional numbers,

The question wants us to show :

Take the reciprocal of square root of 2: ,

Convert 1 by multiplying by this new group, 2a

then we can say



161


1. Similarity

Hydraulic models may be either true or distorted models. True models reproduce

features of the prototype but at a scale - that is they are geometrically similar.

2. Geometric similarity

Geometric similarity exists between model and prototype if the ratio of all

corresponding dimensions in the model and prototype are equal.

where L is the scale factor for length.

For area

All corresponding angles are the same.

3. Kinematic similarity

Kinematic similarity is the similarity of time as well as geometry. It exists between

model and prototype

1. If the paths of moving particles are geometrically similar

2. If the rations of the velocities of particles are similar

Some useful ratios are:

Velocity

Acceleration

Discharge


162

This has the consequence that streamline patterns are the same.

4. Dynamic similarity

Dynamic similarity exists between geometrically and kinematically similar systems if

the ratios of all forces in the model and prototype are the same.

Force ratio

This occurs when the controlling dimensionless group on the right hand side of the

defining equation is the same for model and prototype.

5. Models

When a hydraulic structure is build it undergoes some analysis in the design stage.

Often the structures are too complex for simple mathematical analysis and a hydraulic

model is build. Usually the model is less than full size but it may be greater. The real

structure is known as the prototype. The model is usually built to an exact geometric

scale of the prototype but in some cases - notably river model - this is not possible.

Measurements can be taken from the model and a suitable scaling law applied to

predict the values in the prototype.

To illustrate how these scaling laws can be obtained we will use the relationship for

resistance of a body moving through a fluid.

The resistance, R, is dependent on the following physical properties:

ML-3 u: LT-1 l:(length) L : ML-1T-1

So the defining equation is (R, , u, l, ) = 0

Thus, m = 5, n = 3 so there are 5-3 = 2 groups

For the 1 group

Leading to 1 as

For the 2 group

Leading to 1 as

163

Notice how 1/2 is the Reynolds number. We can call this 2a.

So the defining equation for resistance to motion is

( 1 , 2a ) = 0

We can write

This equation applies whatever the size of the body i.e. it is applicable to a to the

prototype and a geometrically similar model. Thus for the model

and for the prototype

Dividing these two equations gives

At this point we can go no further unless we make some assumptions. One common

assumption is to assume that the Reynolds number is the same for both the model and

prototype i.e.

This assumption then allows the equation following to be written

Which gives this scaling law for resistance force:

164

That the Reynolds numbers were the same was an essential assumption for this

analysis. The consequence of this should be explained.

Substituting this into the scaling law for resistance gives

So the force on the prototype can be predicted from measurement of the force on the

model. But only if the fluid in the model is moving with same Reynolds number as it

would in the prototype. That is to say the Rp can be predicted by

provided that

In this case the model and prototype are dynamically similar.

Formally this occurs when the controlling dimensionless group on the right hand side

of the defining equation is the same for model and prototype. In this case the

controlling dimensionless group is the Reynolds number.

6. Dynamically similar model examples

Example 1

An underwater missile, diameter 2m and length 10m is tested in a water tunnel to

determine the forces acting on the real prototype. A 1/20th scale model is to be used. If

the maximum allowable speed of the prototype missile is 10 m/s, what should be the

speed of the water in the tunnel to achieve dynamic similarity?

165

For dynamic similarity the Reynolds number of the model and prototype must be

equal:

So the model velocity should be

As both the model and prototype are in water then, m = p and m = p so

Note that this is a very high velocity. This is one reason why model tests are not

always done at exactly equal Reynolds numbers. Some relaxation of the equivalence

requirement is often acceptable when the Reynolds number is high. Using a wind

tunnel may have been possible in this example. If this were the case then the

appropriate values of the and ratios need to be used in the above equation.

Example 2

A model aeroplane is built at 1/10 scale and is to be tested in a wind tunnel operating

at a pressure of 20 times atmospheric. The aeroplane will fly at 500km/h. At what

speed should the wind tunnel operate to give dynamic similarity between the model

and prototype? If the drag measure on the model is 337.5 N what will be the drag on

the plane?

From earlier we derived the equation for resistance on a body moving through air:

For dynamic similarity Rem = Rep, so

The value of does not change much with pressure so m = p

The equation of state for an ideal gas is p = RT . As temperature is the same then the

density of the air in the model can be obtained from

166

So the model velocity is found to be

The ratio of forces is found from

So the drag force on the prototype will be

7. Models with free surfaces - rivers, estuaries etc.

When modelling rivers and other fluid with free surfaces the effect of gravity

becomes important and the major governing non-dimensional number becomes the

Froude (Fn) number. The resistance to motion formula above would then be derived

with g as an extra dependent variables to give an extra group. So the defining

equation is:

(R, , u, l, , g ) = 0

From which dimensional analysis gives:

Generally the prototype will have a very large Reynolds number, in which case slight

variation in Re causes little effect on the behaviour of the problem. Unfortunately

models are sometimes so small and the Reynolds numbers are large and the viscous

effects take effect. This situation should be avoided to achieve correct results.

167

Solutions to this problem would be to increase the size of the model - or more difficult

- to change the fluid (i.e. change the viscosity of the fluid) to reduce the Re.

8. Geometric distortion in river models

When river and estuary models are to be built, considerable problems must be

addressed. It is very difficult to choose a suitable scale for the model and to keep

geometric similarity. A model which has a suitable depth of flow will often be far to

big - take up too much floor space. Reducing the size and retaining geometric

similarity can give tiny depth where viscous force come into play. These result in the

following problems:

1. accurate depths and depth changes become very difficult to measure;

2. the bed roughness of the channel becomes impracticably small;

3. laminar flow may result - (turbulent flow is normal in river hydraulics.)

The solution often adopted to overcome these problems is to abandon strict geometric

similarity by having different scales in the horizontal and the vertical. Typical scales

are 1/100 in the vertical and between 1/200 and 1/500 in the horizontal. Good overall

flow patterns and discharge characteristics can be produced by this technique,

however local detail of flow is not well modelled.

In these model the Froude number (u2/d) is used as the dominant non-dimensional

number. Equivalence in Froude numbers can be achieved between model and

prototype even for distorted models. To address the roughness problem artificially

high surface roughness of wire mesh or small blocks is usually used.



Pressure and Manometers

1.1

What will be the (a) the gauge pressure and (b) the absolute pressure of water at depth

12m below the surface? water = 1000 kg/m3, and p atmosphere = 101kN/m2.

[117.72 kN/m2, 218.72 kN/m2]

1.2

At what depth below the surface of oil, relative density 0.8, will produce a pressure of

120 kN/m2? What depth of water is this equivalent to?

[15.3m, 12.2m]

1.3

What would the pressure in kN/m2 be if the equivalent head is measured as 400mm of

(a) mercury =13.6 (b) water ( c) oil specific weight 7.9 kN/m3 (d) a liquid of density



168

520 kg/m3?

[53.4 kN/m2, 3.92 kN/m2, 3.16 kN/m2, 2.04 kN/m2]

1.4

A manometer connected to a pipe indicates a negative gauge pressure of 50mm of

mercury. What is the absolute pressure in the pipe in Newtons per square metre if the

atmospheric pressure is 1 bar?

[93.3 kN/m2]

1.5

What height would a water barometer need to be to measure atmospheric pressure of

1 bar?

[>10.19m]

1.6

An inclined manometer is required to measure an air pressure of 3mm of water to an

accuracy of +/- 3%. The inclined arm is 8mm in diameter and the larger arm has a

diameter of 24mm. The manometric fluid has density 740 kg/m3 and the scale may be

read to +/- 0.5mm.

What is the angle required to ensure the desired accuracy may be achieved?

[7.6]

1.7

Determine the resultant force due to the water acting on the 1m by 2m rectangular

area AB shown in the diagram below.

[43 560 N, 2.37m from O

1.8

Determine the resultant force due to the water acting on the 1.25m by 2.0m triangular

area CD shown in the figure above. The apex of the triangle is at C.

[23.8103N, 2.821m from P]

Forces on submerged surfaces

2.1

Obtain an expression for the depth of the centre of pressure of a plane surface wholly

submerged in a fluid and inclined at an angle to the free surface of the liquid.

169

A horizontal circular pipe, 1.25m diameter, is closed by a butterfly disk which rotates

about a horizontal axis through its centre. Determine the torque which would have to

be applied to the disk spindle to keep the disk closed in a vertical position when there

is a 3m head of fresh water above the axis.

[1176 Nm]

2.2

A dock gate is to be reinforced with three horizontal beams. If the water acts on one

side only, to a depth of 6m, find the positions of the beams measured from the water

surface so that each will carry an equal load. Give the load per meter.

[58 860 N/m, 2.31m, 4.22m, 5.47m]

2.3

The profile of a masonry dam is an arc of a circle, the arc having a radius of 30m and

subtending an angle of 60 at the centre of curvature which lies in the water surface.

Determine (a) the load on the dam in N/m length, (b) the position of the line of action

to this pressure.

[4.28 106 N/m length at depth 19.0m]

2.4

The arch of a bridge over a stream is in the form of a semi-circle of radius 2m. the

bridge width is 4m. Due to a flood the water level is now 1.25m above the crest of the

arch. Calculate (a) the upward force on the underside of the arch, (b) the horizontal

thrust on one half of the arch.

[263.6 kN, 176.6 kN]

2.5

The face of a dam is vertical to a depth of 7.5m below the water surface then slopes at

30 to the vertical. If the depth of water is 17m what is the resultant force per metre

acting on the whole face?

[1563.29 kN]

2.6

A tank with vertical sides is square in plan with 3m long sides. The tank contains oil

of relative density 0.9 to a depth of 2.0m which is floating on water a depth of 1.5m.

Calculate the force on the walls and the height of the centre of pressure from the

bottom of the tank.

[165.54 kN, 1.15m]


3.1







170




metres.

[k = 0.319, 0.0794m]

3.2







[0.816 m3/s]

3.3







[0.23m of water]

3.4



diameter is 0.05m.


the discharge?




[0.0752m, 0.0266m3/s, 0.0118m3/s]

3.5






[0.00195 m3/s]

3.6

The discharge coefficient of a Venturimeter was found to be constant for rates of flow

exceeding a certain value. Show that for this condition the loss of head due to friction

in the convergent parts of the meter can be expressed as KQ2 m where K is a constant

and Q is the rate of flow in cumecs.


171


[K=1060]

3.7







[0.063m]

3.8







[0.0192m3/s, 0.034m3/s]

Tanks emptying

4.1







[1325 seconds]

4.2







4.3






is turned off.



172

orifice.


4.4




to be 0.8.

[1370 seconds]

4.5






[30.7 seconds]

4.6






hour? [3.98m]

Notches and weirs

5.1





from 15cm to 7.5cm.

[1399 seconds]

5.2






[1.24m]

5.3





173




his period.

[0.635]

5.4







[10.94m, 3093seconds]

5.5








[0.755m, 1.218m]

5.6





[0.126m3/s, 0.377m, 0.383m]


6.1






174

6.2





[104.044 kN, 52 29']

6.3






no shear force.)

[338N, 3:1]

6.4



plate.

[2.39kN]

6.5






is 0.075m3.


6.6



[0.018 m3/s]

6.7





6.8






Laminar Pipe Flow

175

7.1

The distribution of velocity, u, in metres/sec with radius r in metres in a smooth bore

tube of 0.025 m bore follows the law, u = 2.5 - kr2. Where k is a constant. The flow is

laminar and the velocity at the pipe surface is zero. The fluid has a coefficient of

viscosity of 0.00027 kg/m s. Determine (a) the rate of flow in m3/s (b) the shearing

force between the fluid and the pipe wall per metre length of pipe.

[6.14x10-4 m3/s, 8.49x10-3 N]

7.2

A liquid whose coefficient of viscosity is m flows below the critical velocity for

laminar flow in a circular pipe of diameter d and with mean velocity u. Show that the

pressure loss in a length of pipe is 32um/d2. Oil of viscosity 0.05 kg/ms flows through a pipe of diameter 0.1m with a velocity of

0.6m/s. Calculate the loss of pressure in a length of 120m.

[11 520 N/m2]

7.3

A plunger of 0.08m diameter and length 0.13m has four small holes of diameter

5/1600 m drilled through in the direction of its length. The plunger is a close fit inside

a cylinder, containing oil, such that no oil is assumed to pass between the plunger and

the cylinder. If the plunger is subjected to a vertical downward force of 45N

(including its own weight) and it is assumed that the upward flow through the four

small holes is laminar, determine the speed of the fall of the plunger. The coefficient

of velocity of the oil is 0.2 kg/ms.

[0.00064 m/s]

7.4

A vertical cylinder of 0.075 metres diameter is mounted concentrically in a drum of

0.076metres internal diameter. Oil fills the space between them to a depth of 0.2m.

The rotque required to rotate the cylinder in the drum is 4Nm when the speed of

rotation is 7.5 revs/sec. Assuming that the end effects are negligible, calculate the

coefficient of viscosity of the oil.

[0.638 kg/ms]

Dimensional analysis

8.1

A stationary sphere in water moving at a velocity of 1.6m/s experiences a drag of 4N.

Another sphere of twice the diameter is placed in a wind tunnel. Find the velocity of

the air and the drag which will give dynamically similar conditions. The ratio of

kinematic viscosities of air and water is 13, and the density of air 1.28 kg/m3.

[10.4m/s 0.865N]

8.2

Explain briefly the use of the Reynolds number in the interpretation of tests on the

flow of liquid in pipes.

Water flows through a 2cm diameter pipe at 1.6m/s. Calculate the Reynolds number

and find also the velocity required to give the same Reynolds number when the pipe is

transporting air. Obtain the ratio of pressure drops in the same length of pipe for both

cases. For the water the kinematic viscosity was 1.3110-6 m2/s and the density was

176

1000 kg/m3. For air those quantities were 15.110-6 m2/s and 1.19kg/m3.

24427, 18.4m/s, 0.157]

8.3

Show that Reynold number, ud/, is non-dimensional. If the discharge Q through an

orifice is a function of the diameter d, the pressure difference p, the density , and the

viscosity , show that Q = Cp1/2d2/1/2 where C is some function of the non-

dimensional group (d1/2d1/2/).

8.4

A cylinder 0.16m in diameter is to be mounted in a stream of water in order to

estimate the force on a tall chimney of 1m diameter which is subject to wind of

33m/s. Calculate (A) the speed of the stream necessary to give dynamic similarity

between the model and chimney, (b) the ratio of forces.

Chimney: = 1.12kg/m3 = 1610-6 kg/ms

Model: = 1000kg/m3 = 810-4 kg/ms

[11.55m/s, 0.057]

8.5

If the resistance to motion, R, of a sphere through a fluid is a function of the density

and viscosity of the fluid, and the radius r and velocity u of the sphere, show that R

is given by

Hence show that if at very low velocities the resistance R is proportional to the

velocity u, then R = kru where k is a dimensionless constant.

A fine granular material of specific gravity 2.5 is in uniform suspension in still water

of depth 3.3m. Regarding the particles as spheres of diameter 0.002cm find how long

it will take for the water to clear. Take k=6 and =0.0013 kg/ms.

[218mins 39.3sec]



Pressure and Manometers

1.1



177

What will be the (a) the gauge pressure and (b) the absolute pressure of water at depth

12m below the surface? water = 1000 kg/m3, and p atmosphere = 101kN/m2.

[117.72 kN/m2, 218.72 kN/m2]

a)

b)

1.2

At what depth below the surface of oil, relative density 0.8, will produce a pressure of

120 kN/m2? What depth of water is this equivalent to?

[15.3m, 12.2m]

a)

b)

1.3

What would the pressure in kN/m2 be if the equivalent head is measured as 400mm of

(a) mercury =13.6 (b) water ( c) oil specific weight 7.9 kN/m3 (d) a liquid of density

520 kg/m3?

[53.4 kN/m2, 3.92 kN/m2, 3.16 kN/m2, 2.04 kN/m2]

a)

178

b)

c)

d)

1.4

A manometer connected to a pipe indicates a negative gauge pressure of 50mm of

mercury. What is the absolute pressure in the pipe in Newtons per square metre is the

atmospheric pressure is 1 bar?

[93.3 kN/m2]

1.5

What height would a water barometer need to be to measure atmospheric pressure?

[>10m]

179

1.6

An inclined manometer is required to measure an air pressure of 3mm of water to an

accuracy of +/- 3%. The inclined arm is 8mm in diameter and the larger arm has a

diameter of 24mm. The manometric fluid has density 740 kg/m3 and the scale may be

read to +/- 0.5mm.

What is the angle required to ensure the desired accuracy may be achieved?

[12 39']

Volume moved from left to right =

180

The head being measured is 3% of 3mm = 0.003x0.03 = 0.00009m

This 3% represents the smallest measurement possible on the manometer, 0.5mm =

0.0005m, giving

1.7

Determine the resultant force due to the water acting on the 1m by 2m rectangular

area AB shown in the diagram below.

[43 560 N, 2.37m from O]

The magnitude of the resultant force on a submerged plane is:

R = pressure at centroid area of surface

This acts at right angle to the surface through the centre of pressure.

181

By the parallel axis theorem (which will be given in an exam), ,

where IGG is the 2nd moment of area about a line through the centroid and can be

found in tables.

For a rectangle

As the wall is vertical, ,

1.8

Determine the resultant force due to the water acting on the 1.25m by 2.0m triangular

area CD shown in the figure above (with question 1.7). The apex of the triangle is at

C.

[43.5103N, 2.821m from P]

For a triangle

Depth to centre of gravity is .

182

Distance from P is

Distance from P to centre of pressure is



Forces on submerged surfaces

2.1

Obtain an expression for the depth of the centre of pressure of a plane surface wholly

submerged in a fluid and inclined at an angle to the free surface of the liquid.

A horizontal circular pipe, 1.25m diameter, is closed by a butterfly disk which rotates

about a horizontal axis through its centre. Determine the torque which would have to

be applied to the disk spindle to keep the disk closed in a vertical position when there

is a 3m head of fresh water above the axis.

[1176 Nm]

Answer:

The question asks what is the moment you have to apply to the spindle to keep the

disc vertical i.e. to keep the valve shut?

So you need to know the resultant force exerted on the disc by the water and the

distance x of this force from the spindle.

We know that the water in the pipe is under a pressure of 3m head of water (to the

spindle)



183

Diagram of the forces on the disc valve, based on an imaginary water surface.

, the depth to the centroid of the disc

h' = depth to the centre of pressure (or line of action of the force)

Calculate the force:

Calculate the line of action of the force, h'.

By the parallel axis theorem 2nd moment of area about O (in the surface)

where IGG is the 2nd moment of area about a line through the centroid

of the disc and IGG = r4/4.

So the distance from the spindle to the line of action of the force is

184

And the moment required to keep the gate shut is

2.2

A dock gate is to be reinforced with three horizontal beams. If the water acts on one

side only, to a depth of 6m, find the positions of the beams measured from the water

surface so that each will carry an equal load. Give the load per meter.

[58 860 N/m, 2.31m, 4.22m, 5.47m]

First of all draw the pressure diagram, as below:

The resultant force per unit length of gate is the area of the pressure diagram. So the

total resultant force is

Alternatively the resultant force is, R = Pressure at centroid Area , (take width of gate

as 1m to give force per m)

This is the resultant force exerted by the gate on the water.

The three beams should carry an equal load, so each beam carries the load f, where

If we take moments from the surface,

185

Taking the first beam, we can draw a pressure diagram for this, (ignoring what is

below),

We know that the resultant force, , so

And the force acts at 2H/3, so this is the position of the 1st beam,

Taking the second beam into consideration, we can draw the following pressure

diagram,

The reaction force is equal to the sum of the forces on each beam, so as before

The reaction force acts at 2H/3, so H=3.27m. Taking moments from the surface,

For the third beam, from before we have,

186

2.3

The profile of a masonry dam is an arc of a circle, the arc having a radius of 30m and

subtending an angle of 60 at the centre of curvature which lies in the water surface.

Determine (a) the load on the dam in N/m length, (b) the position of the line of action

to this pressure.

[4.28 106 N/m length at depth 19.0m]

Draw the dam to help picture the geometry,

Calculate Fv = total weight of fluid above the curved surface (per m length)

Calculate Fh = force on projection of curved surface onto a vertical plane

187

The resultant,

acting at the angle

As this force act normal to the surface, it must act through the centre of radius of the

dam wall. So the depth to the point where the force acts is,

y = 30sin 39.31=19m

2.4

The arch of a bridge over a stream is in the form of a semi-circle of radius 2m. the

bridge width is 4m. Due to a flood the water level is now 1.25m above the crest of the

arch. Calculate (a) the upward force on the underside of the arch, (b) the horizontal

thrust on one half of the arch.

[263.6 kN, 176.6 kN]

The bridge and water level can be drawn as:

1. The upward force on the arch = weight of (imaginary) water above the arch.

188

b)

The horizontal force on half of the arch, is equal to the force on the projection of the

curved surface onto a vertical plane.

2.5

The face of a dam is vertical to a depth of 7.5m below the water surface then slopes at

30 to the vertical. If the depth of water is 17m what is the resultant force per metre

acting on the whole face?

[1563.29 kN]

h2 = 17.0 m, so h1 = 17.0 - 7.5 = 9.5 . x = 9.5/tan 60 = 5.485 m.

Vertical force = weight of water above the surface,

The horizontal force = force on the projection of the surface on to a vertical plane.

189

The resultant force is

And acts at the angle

2.6

A tank with vertical sides is square in plan with 3m long sides. The tank contains oil

of relative density 0.9 to a depth of 2.0m which is floating on water a depth of 1.5m.

Calculate the force on the walls and the height of the centre of pressure from the

bottom of the tank.

[165.54 kN, 1.15m]

Consider one wall of the tank. Draw the pressure diagram:

density of oil oil = 0.9water = 900 kg/m3.

Force per unit length, F = area under the graph = sum of the three areas = f1 + f2 + f3

190

To find the position of the resultant force F, we take moments from any point. We

will take moments about the surface.




3.1










metres.

[k = 0.319, 0.0794m]



191

Part i)

Taking the datum at B, the Bernoulli equation becomes:

By continuity: Q = uAAA = uBAB

giving

Part ii)

3.2



192





[0.816 m3/s]

What we know from the question:

Calculate Q.

For the manometer:

For the Venturimeter

193

Combining (1) and (2)

3.3







[0.23m of water]

For the manometer:

For the Venturimeter

194

Combining (1) and (2)

but at 1. From the question

Substitute in (3)

3.4



diameter is 0.05m.


the discharge?




[0.0752m, 0.0266m3/s, 0.0118m3/s]

195

From the question:

Apply Bernoulli:

If we take the datum through the orifice:

Between 1 and 2

Between 1 and 3

196

If the mouth piece has been removed,

3.5






[0.00195 m3/s]

From the question

197

Apply Bernoulli,

Take atmospheric pressure as 0,

3.6

The discharge of a Venturimeter was found to be constant for rates of flow exceeding

a certain value. Show that for this condition the loss of head due to friction in the

convergent parts of the meter can be expressed as KQ2 m where K is a constant and Q

is the rate of flow in cumecs.



[K=1060]

3.7







[0.063m]

From the question:

198

Friction loss, from the question:

Apply Bernoulli:

3.8







[0.0192m3/s, 0.034m3/s]

199

From the question:

Apply Bernoulli:

1.

By continuity:

200

b)



Tank emptying

4.1







[1325 seconds]



201

From the question: H = 4m a = (0.65/2)2 = 0.33m2

In time t the level in the reservoir falls h, so

Integrating give the total time for levels to fall from h1 to h2.

As the surface area changes with height, we must express A in terms of h.

A = r2

But r varies with h.

It varies linearly from the surface at H = 4m, r = 25m, at a gradient of tan-1 = 1/5.

r = x + 5h

25 = x + 5(4)

x = 5

so A = ( 5 + 5h )2 = ( 25 + 25h2 + 50h )

202

Substituting in the integral equation gives

From the question, h1 = 4m h2 = 2m, so

4.2







The question tell us ao = 0.224m2, Cd = 0.6

Apply Bernoulli from the tank surface to the vena contracta at the orifice:

203

p1 = p2 and u1 = 0.

We need Q in terms of the height h measured above the orifice.

And we can write an equation for the discharge in terms of the surface height change:

Integrating give the total time for levels to fall from h1 to h2.

a) For the first 1m depth, A = 8 x 32 = 256, whatever the h.

So, for the first period of time:

b) now we need to find out how long it will take to empty the rest.

We need the area A, in terms of h.

204

So

Total time for emptying is,

T = 363 + 299 = 662 sec

4.3






is turned off.



orifice.


From the question: Qin = 0.0095 m3/s, do=0.05m, Cd =0.6

Apply Bernoulli from the water surface (1) to the orifice (2),

205

p1 = p2 and u1 = 0. .

With the datum the bottom of the cylinder, z1 = h, z2 = 0


For the level in the tank to remain constant:

inflow = out flow

Qin = Qout

(b) Write the equation for the discharge in terms of the surface height change:

Integrating between h1 and h2, to give the time to change surface level

h1 = 3 and h2 = 1 so

T = 881 sec

1. Qin changed to Qin = 0.02 m3/s

From (1) we have . The question asks for the rate of surface rise

when h = 1.7m.

206

i.e.

The rate of increase in volume is:

As Q = Area x Velocity, the rate of rise in surface is

4.4




to be 0.8.

[1370 seconds]

From the question W = 10m, D = 10m do = 0.08m Cd = 0.8

Apply Bernoulli from the water surface (1) to the orifice (2),

p1 = p2 and u1 = 0. .

With the datum the bottom of the cylinder, z1 = h, z2 = 0


207

Write the equation for the discharge in terms of the surface height change:


But we need A in terms of h.

Surface area A = 10L, so need L in terms of h

Substitute this into the integral term,

208

4.5






[30.7 seconds]

by continuity,

209

defining, h = h1 - h2

Substituting this in (1) to eliminate h2

From the Bernoulli equation we can derive this expression for discharge through the

submerged orifice:

So

Integrating

4.6






210

hour? [3.98m]

From the question A = 60 000 m2, Q = 0.678 h 3/2



From the question T = 3600 sec and h1 = 0.6m

Total depth = 3.4 + 0.58 = 3.98m



Notches and weirs



211

5.1





from 15cm to 7.5cm.

[1399 seconds]

From your notes you can derive:

For this weir the equation simplifies to



h1 = 0.15m, h2 = 0.075m

212

5.2






[1.24m]

From your notes you can derive:

From the question:

Q1 = 0.2 m3/s, h1 = x

Q2 = 1.0 m3/s, h2 = x + 0.384

where x is the height above the weir at normal flow.

So we have two situations:

From (1) we get an expression for b in terms of x

Substituting this in (2) gives,

213

So the weir breadth is

5.3








his period.

[0.635]

The proof for is in the notes.

From the question:

A = 10m2 = 90 h1 = 0.08m h2 = 0.07m T = 43.5sec

So

Q = 2.36 Cd h5/2



214

5.4







[10.94m, 3093seconds]

From the question:

A = 56000 m2 Q = 1.77 B H 3/2 Qmax = 9 m3/s

a) Find B for H = 0.6

9 = 1.77 B 0.63/2

B = 10.94m

b) Write the equation for the discharge in terms of the surface height change:


215

5.5








[0.755m, 1.218m]

Derive this formula from the notes:

From the question:

= 90 Cd 0.58 Q = 0.3 m3/s, depth of water, Z = 0.3m

giving the weir equation:

a) As H is the height above the bottom of the V, the depth of water = Z = D + H,

where D is the height of the bottom of the V from the base of the channel. So

1. Find Z when Q = 0.2 m3/s

5.6





[0.126m3/s, 0.377m, 0.383m]

Proof of the v-notch weir equation is in the notes.

From the question:

H = 0.38m = 45 Cd = 0.6

216

The weir equation becomes:

Q+2% = 0.129 m3/s

Q-2% = 0.124 m3/s




6.1






From the question:



217

Calculate the total force using the momentum equation:

Body force and pressure force are 0.

So force on vane:

6.2





[104.044 kN, 52 29']

From the question:

218

Calculate total force.

Calculate the pressure force

p1 = p2 = p = hg = 3010009.81 = 294.3 kN/m2

There is no body force in the x or y directions.

These forces act on the fluid

The resultant force on the fluid is

6.3





219


no shear force.)

[338N, 3:1]

From the question a2 = a3 =2x10-3 m2 u = 15 m/s

Apply Bernoulli,

Change in height is negligible so z1 = z2 = z3 and pressure is always atmospheric p1=

p2 = p3 =0. So

u1= u2 = u3 =15 m/s

By continuity Q1= Q2 + Q3

u1a1 = u2a2 + u3a3

so a1 = a2 + a3

Put the axes normal to the plate, as we know that the resultant force is normal to the

plate.

Q1 = a1u = 210-315 = 0.03

Q1 = (a2 + a3) u

Q2 = a2u

220

Q3 = (a1 - a2)u

Calculate total force.

Component in direction of jet = 390 sin 60 = 338 N

As there is no force parallel to the plate Fty = 0

Thus 3/4 of the jet goes up, 1/4 down

6.4



plate.

[2.39kN]

221

From the question, djet = 0.075m u1=25m/s Q = 25(0.075/2)2 = 0.11 m3/s

Force normal to plate is

FTx = Q( 0 - u1x )

FTx = 10000.11 ( 0 - 25 cos 30 ) = 2.39 kN

6.5






is 0.075m3.


1&2 Draw the control volume and the axis system

p1 = 100 kN/m2, Q = 0.3 m3/s = 45

d1 = 0.15 m d2 = 0.3 m

A1 = 0.177 m2 A2 = 0.0707 m2


in the x direction

222

by continuity , so



We know pressure at the inlet but not at the outlet.

we can use Bernoulli to calculate this unknown pressure.

where hf is the friction loss

In the question it says this can be ignored, hf=0

The height of the pipe at the outlet is 1m above the inlet.

Taking the inlet level as the datum:

z1 = 0 z2 = 1m

So the Bernoulli equation becomes:

223


The only body force is the force due to gravity. That is the weight acting in the y

direction.

There are no body forces in the x direction,



224


The force on the bend is the same magnitude but in the opposite direction

6.6



[0.018 m3/s]

From the question, djet = 0.025m FTx = 650 N

225

Force normal to plate is

FTx = Q( 0 - u1x )

650 = 1000Q ( 0 - u )

Q = au = (d2/4)u

650 = -1000au2 = -1000Q2/a

650 = -1000Q2/(0.0252/4)

Q = 0.018m3/s

6.7





From the question:

Calculate the total force using the momentum equation:

226

Body force and pressure force are 0.

So force on vane:

6.8






1&2 Draw the control volume and the axis system

p1 = 1.45105 N/m2, Q = 0.45 m3/s = 45

d1 = 0.6 m d2 = 0.3 m

A1 = 0.283 m2 A2 = 0.0707 m2

227


in the x direction

by continuity , so



We know pressure at the inlet but not at the outlet.

we can use Bernoulli to calculate this unknown pressure.

where hf is the friction loss

In the question it says this can be ignored, hf=0

228

Assume the pipe to be horizontal

z1 = z2

So the Bernoulli equation becomes:


The only body force is the force due to gravity.

There are no body forces in the x or y directions,



229


The force on the bend is the same magnitude but in the opposite direction



Laminar flow in pipes examples.

7.1

The distribution of velocity, u, in metres/sec with radius r in metres in a smooth bore

tube of 0.025 m bore follows the law, u = 2.5 - kr2. Where k is a constant. The flow is

laminar and the velocity at the pipe surface is zero. The fluid has a coefficient of

viscosity of 0.00027 kg/m s. Determine (a) the rate of flow in m3/s (b) the shearing

force between the fluid and the pipe wall per metre length of pipe.

[6.14x10-4 m3/s, 8.49x10-3 N]

The velocity at distance r from the centre is given in the question:

u = 2.5 - kr2

Also we know: = 0.00027 kg/ms 2r = 0.025m

We can find k from the boundary conditions:



230

when r = 0.0125, u = 0.0 (boundary of the pipe)

0.0 = 2.5 - k0.01252

k = 16000

u = 2.5 - 1600 r2

a)

Following along similar lines to the derivation seen in the lecture notes, we

can calculate the flow Q through a small annulus r:

b)

The shear force is given by F = (2r)

From Newtons law of viscosity

7.2

A liquid whose coefficient of viscosity is m flows below the critical velocity for

laminar flow in a circular pipe of diameter d and with mean velocity u. Show that the

pressure loss in a length of pipe is 32um/d2. Oil of viscosity 0.05 kg/ms flows through a pipe of diameter 0.1m with a velocity of

0.6m/s. Calculate the loss of pressure in a length of 120m.

[11 520 N/m2]

See the proof in the lecture notes for

Consider a cylinder of fluid, length L, radius r, flowing steadily in the centre of a pipe

231

The fluid is in equilibrium, shearing forces equal the pressure forces.

Newtons law of viscosity ,

We are measuring from the pipe centre, so

Giving:

In an integral form this gives an expression for velocity,

The value of velocity at a point distance r from the centre

At r = 0, (the centre of the pipe), u = umax, at r = R (the pipe wall) u = 0;

At a point r from the pipe centre when the flow is laminar:

232

The flow in an annulus of thickness r

So the discharge can be written

To get pressure loss in terms of the velocity of the flow, use the mean velocity:

1. From the question = 0.05 kg/ms d = 0.1m

u = 0.6 m/s L = 120.0m

7.3

A plunger of 0.08m diameter and length 0.13m has four small holes of diameter

5/1600 m drilled through in the direction of its length. The plunger is a close fit inside

a cylinder, containing oil, such that no oil is assumed to pass between the plunger and

the cylinder. If the plunger is subjected to a vertical downward force of 45N

(including its own weight) and it is assumed that the upward flow through the four

small holes is laminar, determine the speed of the fall of the plunger. The coefficient

233

of velocity of the oil is 0.2 kg/ms.

[0.00064 m/s]

Flow through each tube given by Hagen-Poiseuille equation

There are 4 of these so total flow is

Force = pressure area

Flow up through piston = flow displaced by moving piston

Q = Avpiston

3.2410-6 = 0.042vpiston

vpiston = 0.00064 m/s

7.4

A vertical cylinder of 0.075 metres diameter is mounted concentrically in a drum of

0.076metres internal diameter. Oil fills the space between them to a depth of 0.2m.

The rotque required to rotate the cylinder in the drum is 4Nm when the speed of

rotation is 7.5 revs/sec. Assuming that the end effects are negligible, calculate the

coefficient of viscosity of the oil.

[0.638 kg/ms]

From the question r-1 = 0.076/2 r2 = 0.075/2 Torque = 4Nm, L = 0.2m

The velocity of the edge of the cylinder is:

ucyl = 7.5 2r = 7.520.0375 = 1.767 m/s

udrum = 0.0

Torque needed to rotate cylinder

234

Distance between cylinder and drum = r1 - r2 = 0.038 - 0.0375 = 0.005m

Using Newtons law of viscosity:



Dimensional analysis

8.1

A stationary sphere in water moving at a velocity of 1.6m/s experiences a drag of 4N.

Another sphere of twice the diameter is placed in a wind tunnel. Find the velocity of

the air and the drag which will give dynamically similar conditions. The ratio of

kinematic viscosities of air and water is 13, and the density of air 1.28 kg/m3.

[10.4m/s 0.865N]

Draw up the table of values you have for each variable:

variable water air

u 1.6m/s uair

Drag 4N Dair

13

1000 kg/m3 1.28 kg/m3

d d 2d

Kinematic viscosity is dynamic viscosity over density =

The Reynolds number =



235

Choose the three recurring (governing) variables; u, d,

From Buckinghams theorem we have m-n = 5 - 3 = 2 non-dimensional groups.

As each group is dimensionless then considering the dimensions, for the first group,

1:

(note D is a force with dimensions MLT-2)

M] 0 = c1 + 1

c1 = -1

L] 0 = a1 + b1 - 3c1 + 1

-4 = a1 + b1

T] 0 = -a1 - 2

a1 = - 2

b1 = -2


M] 0 = c2

L] 0 = a2 + b2 - 3c2 + 2

-2 = a2 + b2

T] 0 = -a2 - 1

236

a2 = -1

b2 = -1

So the physical situation is described by this function of nondimensional numbers,

For dynamic similarity these non-dimensional numbers are the same for the both the

sphere in water and in the wind tunnel i.e.

For 1

For 2

8.2

Explain briefly the use of the Reynolds number in the interpretation of tests on the

flow of liquid in pipes.

Water flows through a 2cm diameter pipe at 1.6m/s. Calculate the Reynolds number

and find also the velocity required to give the same Reynolds number when the pipe is

transporting air. Obtain the ratio of pressure drops in the same length of pipe for both

cases. For the water the kinematic viscosity was 1.3110-6 m2/s and the density was

1000 kg/m3. For air those quantities were 15.110-6 m2/s and 1.19kg/m3.

[24427, 18.4m/s, 0.157]

237


variable water air

u 1.6m/s uair

p pwater pair

1000 kg/m3 1.19kg/m3

ms ms

1000 kg/m3 1.28 kg/m3

d 0.02m 0.02m



Reynolds number when carrying water:

To calculate Reair we know,

To obtain the ratio of pressure drops we must obtain an expression for the pressure

drop in terms of governing variables.

Choose the three recurring (governing) variables; u, d,



1:

238

M] 0 = c1

L] 0 = a1 + b1 - 3c1 + 2

-2 = a1 + b1

T] 0 = -a1 - 1

a1 = -1

b1 = -1



M] 0 = c2 + 1

c2 = -1

L] 0 = a2 + b2 - 3c2 - 1

-2 = a2 + b2

T] 0 = -a2 - 2

a2 = - 2

b2 = 0


For dynamic similarity these non-dimensional numbers are the same for the both

water and air in the pipe.

239

We are interested in the relationship involving the pressure i.e. 2

8.3

Show that Reynold number, ud/, is non-dimensional. If the discharge Q through an

orifice is a function of the diameter d, the pressure difference p, the density , and the

viscosity , show that Q = Cp1/2d2/1/2 where C is some function of the non-

dimensional group (d1/2p1/2/).


The dimensions of these following variables are

ML-3

u LT-1

d L

ML-1T-1

Re = ML-3 LT-1L(ML-1T-1)-1 = ML-3 LT-1 L M-1LT = 1

i.e. Re is dimensionless.

We are told from the question that there are 5 variables involved in the problem: d, p,

, and Q.

Choose the three recurring (governing) variables; Q, d,


240


1:

M] 0 = c1 + 1

c1 = -1

L] 0 = 3a1 + b1 - 3c1 - 1

-2 = 3a1 + b1

T] 0 = -a1 - 1

a1 = -1

b1 = 1



M] 0 = c2 + 1

c2 = -1

L] 0 = 3a2 + b2 - 3c2 - 1

-2 = 3a2 + b2

T] 0 = -a2 - 2

a2 = - 2

b2 = 4

241

So the physical situation is described by this function of non-dimensional numbers,

The question wants us to show :

Take the reciprocal of square root of 2: ,

Convert 1 by multiplying by this number

then we can say

8.4

A cylinder 0.16m in diameter is to be mounted in a stream of water in order to

estimate the force on a tall chimney of 1m diameter which is subject to wind of

33m/s. Calculate (A) the speed of the stream necessary to give dynamic similarity

between the model and chimney, (b) the ratio of forces.

Chimney: = 1.12kg/m3 = 1610-6 kg/ms

Model: = 1000kg/m3 = 810-4 kg/ms

[11.55m/s, 0.057]


242

variable water air

u uwater 33m/s

F Fwater Fair

1000 kg/m3 1.12kg/m3

kgms kg/ms

d 0.16m 1m



For dynamic similarity:

To obtain the ratio of forces we must obtain an expression for the force in terms of

governing variables.

Choose the three recurring (governing) variables; u, d, F,



1:

M] 0 = c1 + 1

c1 = -1

L] 0 = a1 + b1 - 3c1 - 1

-2 = a1 + b1

243

T] 0 = -a1 - 1

a1 = -1

b1 = -1

i.e. the (inverse of) Reynolds number


M] 0 = c2 + 1

c2 = -1

L] 0 = a2 + b2 - 3c2 - 1

-3 = a2 + b2

T] 0 = -a2 - 2

a2 = - 2

b2 = -1


For dynamic similarity these non-dimensional numbers are the same for the both

water and air in the pipe.

To find the ratio of forces for the different fluids use 2

244

8.5

If the resistance to motion, R, of a sphere through a fluid is a function of the density

and viscosity of the fluid, and the radius r and velocity u of the sphere, show that R

is given by

Hence show that if at very low velocities the resistance R is proportional to the

velocity u, then R = kru where k is a dimensionless constant.

A fine granular material of specific gravity 2.5 is in uniform suspension in still water

of depth 3.3m. Regarding the particles as spheres of diameter 0.002cm find how long

it will take for the water to clear. Take k=6 and =0.0013 kg/ms.

[218mins 39.3sec]

Choose the three recurring (governing) variables; u, r, R,



1:

M] 0 = c1 + 1

c1 = -1

L] 0 = a1 + b1 - 3c1 - 1

245

-2 = a1 + b1

T] 0 = -a1 - 1

a1 = -1

b1 = -1

i.e. the (inverse of) Reynolds number


M] 0 = c2 + 1

c2 = -1

L] 0 = a2 + b2 - 3c2 - 1

-3 = a2 + b2

T] 0 = -a2 - 2

a2 = - 2

b2 = -1


or

246

he question asks us to show or

Multiply the LHS by the square of the RHS: (i.e. 2(1/12) )

So

The question tells us that R is proportional to u so the function f must be a constant, k

The water will clear when the particle moving from the water surface reaches the

bottom.

At terminal velocity there is no acceleration - the force R = mg - upthrust.

From the question:

= 2.5 so = 2500kg/m3 = 0.0013 kg/ms k = 6

r = 0.00001m depth = 3.3m

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